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\section*{Handout 5 (Grammars \& Parser)}

While regular expressions are very useful for lexing and for

recognising many patterns in strings (like email addresses),

they have their limitations. For example there is no regular

expression that can recognise the language $a^nb^n$. Another

example for which there exists no regular expression is the

language of well-parenthesised expressions. In languages like

Lisp, which use parentheses rather extensively, it might be of

interest to know whether the following two expressions are

well-parenthesised or not (the left one is, the right one is not):

\begin{center}

$(((()()))())$  \hspace{10mm} $(((()()))()))$

\end{center}

\noindent Not being able to solve such recognition problems is

a serious limitation. In order to solve such recognition

problems, we need more powerful techniques than regular

expressions. We will in particular look at \emph{context-free

languages}. They include the regular languages as the picture

below about language classes shows:

\begin{center}

\begin{tikzpicture}

[rect/.style={draw=black!50, 

 top color=white,bottom color=black!20, 

 rectangle, very thick, rounded corners}]

\draw (0,0) node [rect, text depth=30mm, text width=46mm] {\small all languages};

\draw (0,-0.4) node [rect, text depth=20mm, text width=44mm] {\small decidable languages};

\draw (0,-0.65) node [rect, text depth=13mm] {\small context sensitive languages};

\draw (0,-0.84) node [rect, text depth=7mm, text width=35mm] {\small context-free languages};

\draw (0,-1.05) node [rect] {\small regular languages};

\end{tikzpicture}

\end{center}

\noindent Each ``bubble'' stands for sets of languages (remember

languages are sets of strings). As indicated the set of regular

languages are fully included inside the context-free languages,

meaning every regular language is also context-free, but not vice

versa. Below I will let you think for example what the context-free

grammar is for the language corresponding to the regular expression

$(aaa)^*a$.

Because of their convenience, context-free languages play an important

role in `day-to-day' text processing and in programming

languages. Context-free in this setting means that ``words'' have one

meaning only ansd this meaning is independend from in which context

the ``words'' appear. For example ambiguity issues like

\begin{center}

\tt Time flies like an arrow; fruit flies like a banana.

\end{center}  

\noindent

from natural languages were the meaning of \emph{flies} depend on the

surounding \emph{context} are avoided as much as possible.

Context-free languages are usually specified by grammars. For example

a grammar for well-parenthesised expressions can be given as follows:

\begin{plstx}[margin=3cm]

: \meta{P} ::=  ( \cdot  \meta{P} \cdot ) \cdot \meta{P}

             | \epsilon\\ 

\end{plstx}

\noindent 

or a grammar for recognising strings consisting of ones is

\begin{plstx}[margin=3cm]

: \meta{O} ::= 1 \cdot  \meta{O} 

             | 1\\

\end{plstx}

In general grammars consist of finitely many rules built up

from \emph{terminal symbols} (usually lower-case letters) and

\emph{non-terminal symbols} (upper-case letters written in

bold like \meta{A}, \meta{N} and so on). Rules have

the shape

\begin{plstx}[margin=3cm]

: \meta{NT} ::= rhs\\

\end{plstx}

\noindent where on the left-hand side is a single non-terminal

and on the right a string consisting of both terminals and

non-terminals including the $\epsilon$-symbol for indicating

the empty string. We use the convention to separate components

on the right hand-side by using the $\cdot$ symbol, as in the

grammar for well-parenthesised expressions. We also use the

convention to use $|$ as a shorthand notation for several

rules. For example

\begin{plstx}[margin=3cm]

: \meta{NT} ::= rhs_1

   | rhs_2\\

\end{plstx}

\noindent means that the non-terminal \meta{NT} can be replaced by

either $\textit{rhs}_1$ or $\textit{rhs}_2$. If there are more

than one non-terminal on the left-hand side of the rules, then

we need to indicate what is the \emph{starting} symbol of the

grammar. For example the grammar for arithmetic expressions

can be given as follows

\begin{plstx}[margin=3cm,one per line]

\mbox{\rm (1)}: \meta{E} ::= \meta{N}\\

\mbox{\rm (2)}: \meta{E} ::= \meta{E} \cdot + \cdot \meta{E}\\

\mbox{\rm (3)}: \meta{E} ::= \meta{E} \cdot - \cdot \meta{E}\\

\mbox{\rm (4)}: \meta{E} ::= \meta{E} \cdot * \cdot \meta{E}\\

\mbox{\rm (5)}: \meta{E} ::= ( \cdot \meta{E} \cdot )\\

\mbox{\rm (6\ldots)}: \meta{N} ::= \meta{N} \cdot \meta{N} 

  \mid 0 \mid 1 \mid \ldots \mid 9\\

\end{plstx}

\noindent where \meta{E} is the starting symbol. A

\emph{derivation} for a grammar starts with the starting

symbol of the grammar and in each step replaces one

non-terminal by a right-hand side of a rule. A derivation ends

with a string in which only terminal symbols are left. For

example a derivation for the string $(1 + 2) + 3$ is as

follows:

\begin{center}

\begin{tabular}{lll@{\hspace{2cm}}l}

\meta{E} & $\rightarrow$ & $\meta{E}+\meta{E}$          & by (2)\\

       & $\rightarrow$ & $(\meta{E})+\meta{E}$     & by (5)\\

       & $\rightarrow$ & $(\meta{E}+\meta{E})+\meta{E}$   & by (2)\\

       & $\rightarrow$ & $(\meta{E}+\meta{E})+\meta{N}$   & by (1)\\

       & $\rightarrow$ & $(\meta{E}+\meta{E})+3$   & by (6\dots)\\

       & $\rightarrow$ & $(\meta{N}+\meta{E})+3$   & by (1)\\

       & $\rightarrow^+$ & $(1+2)+3$ & by (1, 6\ldots)\\

\end{tabular} 

\end{center}

\noindent where on the right it is indicated which 

grammar rule has been applied. In the last step we

merged several steps into one.

The \emph{language} of a context-free grammar $G$

with start symbol $S$ is defined as the set of strings

derivable by a derivation, that is

\begin{center}

$\{c_1\ldots c_n \;|\; S \rightarrow^* c_1\ldots c_n \;\;\text{with all} \; c_i \;\text{being non-terminals}\}$

\end{center}

\noindent

A \emph{parse-tree} encodes how a string is derived with the starting symbol on 

top and each non-terminal containing a subtree for how it is replaced in a derivation.

The parse tree for the string $(1 + 23)+4$ is as follows:

\begin{center}

\begin{tikzpicture}[level distance=8mm, black]

  \node {$E$}

    child {node {$E$} 

       child {node {$($}}

       child {node {$E$}       

         child {node {$E$} child {node {$N$} child {node {$1$}}}}

         child {node {$+$}}

         child {node {$E$} 

            child {node {$N$} child {node {$2$}}}

            child {node {$N$} child {node {$3$}}}

            } 

        }

       child {node {$)$}}

     }

     child {node {$+$}}

     child {node {$E$}

        child {node {$N$} child {node {$4$}}}

     };

\end{tikzpicture}

\end{center}

\noindent We are often interested in these parse-trees since

they encode the structure of how a string is derived by a

grammar. Before we come to the problem of constructing such

parse-trees, we need to consider the following two properties

of grammars. A grammar is \emph{left-recursive} if there is a

derivation starting from a non-terminal, say $NT$ which leads

to a string which again starts with $NT$. This means a

derivation of the form.

\begin{center}

$NT \rightarrow \ldots \rightarrow NT \cdot \ldots$

\end{center}

\noindent It can be easily seen that the grammar above for

arithmetic expressions is left-recursive: for example the

rules $E \rightarrow E\cdot + \cdot E$ and $N \rightarrow

N\cdot N$ show that this grammar is left-recursive. But note

that left-recursiveness can involve more than one step in the

derivation. The problem with left-recursive grammars is that

some algorithms cannot cope with them: they fall into a loop.

Fortunately every left-recursive grammar can be transformed

into one that is not left-recursive, although this

transformation might make the grammar less ``human-readable''.

For example if we want to give a non-left-recursive grammar

for numbers we might specify

\begin{center}

$N \;\;\rightarrow\;\; 0\;|\;\ldots\;|\;9\;|\;1\cdot N\;|\;2\cdot N\;|\;\ldots\;|\;9\cdot N$

\end{center}

\noindent Using this grammar we can still derive every number

string, but we will never be able to derive a string of the

form $N \to \ldots \to N \cdot \ldots$.

The other property we have to watch out for is when a grammar

is \emph{ambiguous}. A grammar is said to be ambiguous if

there are two parse-trees for one string. Again the grammar

for arithmetic expressions shown above is ambiguous. While the

shown parse tree for the string $(1 + 23) + 4$ is unique, this

is not the case in general. For example there are two parse

trees for the string $1 + 2 + 3$, namely

\begin{center}

\begin{tabular}{c@{\hspace{10mm}}c}

\begin{tikzpicture}[level distance=8mm, black]

  \node {$E$}

    child {node {$E$} child {node {$N$} child {node {$1$}}}}

    child {node {$+$}}

    child {node {$E$}

       child {node {$E$} child {node {$N$} child {node {$2$}}}}

       child {node {$+$}}

       child {node {$E$} child {node {$N$} child {node {$3$}}}}

    }

    ;

\end{tikzpicture} 

&

\begin{tikzpicture}[level distance=8mm, black]

  \node {$E$}

    child {node {$E$}

       child {node {$E$} child {node {$N$} child {node {$1$}}}}

       child {node {$+$}}

       child {node {$E$} child {node {$N$} child {node {$2$}}}} 

    }

    child {node {$+$}}

    child {node {$E$} child {node {$N$} child {node {$3$}}}}

    ;

\end{tikzpicture}

\end{tabular} 

\end{center}

\noindent In particular in programming languages we will try

to avoid ambiguous grammars because two different parse-trees

for a string mean a program can be interpreted in two

different ways. In such cases we have to somehow make sure the

two different ways do not matter, or disambiguate the grammar

in some other way (for example making the $+$

left-associative). Unfortunately already the problem of

deciding whether a grammar is ambiguous or not is in general

undecidable. But in simple instance (the ones we deal in this

module) one can usually see when a grammar is ambiguous.

\subsection*{Parser Combinators}

Let us now turn to the problem of generating a parse-tree for

a grammar and string. In what follows we explain \emph{parser

combinators}, because they are easy to implement and closely

resemble grammar rules. Imagine that a grammar describes the

strings of natural numbers, such as the grammar $N$ shown

above. For all such strings we want to generate the

parse-trees or later on we actually want to extract the

meaning of these strings, that is the concrete integers

``behind'' these strings. In Scala the parser combinators will

be functions of type

\begin{center}

\texttt{I $\Rightarrow$ Set[(T, I)]}

\end{center}

\noindent that is they take as input something of type

\texttt{I}, typically a list of tokens or a string, and return

a set of pairs. The first component of these pairs corresponds

to what the parser combinator was able to process from the

input and the second is the unprocessed part of the input. As

we shall see shortly, a parser combinator might return more

than one such pair, with the idea that there are potentially

several ways how to interpret the input. As a concrete

example, consider the case where the input is of type string,

say the string

\begin{center}

\tt\Grid{iffoo\VS testbar}

\end{center}

\noindent We might have a parser combinator which tries to

interpret this string as a keyword (\texttt{if}) or an

identifier (\texttt{iffoo}). Then the output will be the set

\begin{center}

$\left\{ \left(\texttt{\Grid{if}}\,,\, \texttt{\Grid{foo\VS testbar}}\right), 

           \left(\texttt{\Grid{iffoo}}\,,\, \texttt{\Grid{\VS testbar}}\right) \right\}$

\end{center}

\noindent where the first pair means the parser could

recognise \texttt{if} from the input and leaves the rest as

`unprocessed' as the second component of the pair; in the

other case it could recognise \texttt{iffoo} and leaves

\texttt{\VS testbar} as unprocessed. If the parser cannot

recognise anything from the input then parser combinators just

return the empty set $\{\}$. This will indicate

something ``went wrong''.

The main attraction is that we can easily build parser combinators out of smaller components

following very closely the structure of a grammar. In order to implement this in an object

oriented programming language, like Scala, we need to specify an abstract class for parser 

combinators. This abstract class requires the implementation of the function

\texttt{parse} taking an argument of type \texttt{I} and returns a set of type  

\mbox{\texttt{Set[(T, I)]}}.

\begin{center}

\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]

abstract class Parser[I, T] {

  def parse(ts: I): Set[(T, I)]

  def parse_all(ts: I): Set[T] =

    for ((head, tail) <- parse(ts); if (tail.isEmpty)) 

      yield head

}

\end{lstlisting}

\end{center}

\noindent

From the function \texttt{parse} we can then ``centrally'' derive the function \texttt{parse\_all},

which just filters out all pairs whose second component is not empty (that is has still some

unprocessed part). The reason is that at the end of parsing we are only interested in the

results where all the input has been consumed and no unprocessed part is left.

One of the simplest parser combinators recognises just a character, say $c$, 

from the beginning of strings. Its behaviour is as follows:

\begin{itemize}

\item if the head of the input string starts with a $c$, it returns 

	the set $\{(c, \textit{tail of}\; s)\}$

\item otherwise it returns the empty set $\varnothing$	

\end{itemize}

\noindent

The input type of this simple parser combinator for characters is

\texttt{String} and the output type \mbox{\texttt{Set[(Char, String)]}}. 

The code in Scala is as follows:

\begin{center}

\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]

case class CharParser(c: Char) extends Parser[String, Char] {

  def parse(sb: String) = 

    if (sb.head == c) Set((c, sb.tail)) else Set()

}

\end{lstlisting}

\end{center}

\noindent

The \texttt{parse} function tests whether the first character of the 

input string \texttt{sb} is equal to \texttt{c}. If yes, then it splits the

string into the recognised part \texttt{c} and the unprocessed part

\texttt{sb.tail}. In case \texttt{sb} does not start with \texttt{c} then

the parser returns the empty set (in Scala \texttt{Set()}).

More interesting are the parser combinators that build larger parsers

out of smaller component parsers. For example the alternative 

parser combinator is as follows.

\begin{center}

\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]

class AltParser[I, T]

       (p: => Parser[I, T], 

        q: => Parser[I, T]) extends Parser[I, T] {

  def parse(sb: I) = p.parse(sb) ++ q.parse(sb)

}

\end{lstlisting}

\end{center}

\noindent

The types of this parser combinator are polymorphic (we just have \texttt{I}

for the input type, and \texttt{T} for the output type). The alternative parser

builds a new parser out of two existing parser combinator \texttt{p} and \texttt{q}.

Both need to be able to process input of type \texttt{I} and return the same

output type \texttt{Set[(T, I)]}. (There is an interesting detail of Scala, namely the 

\texttt{=>} in front of the types of \texttt{p} and \texttt{q}. They will prevent the

evaluation of the arguments before they are used. This is often called 

\emph{lazy evaluation} of the arguments.) The alternative parser should run

the input with the first parser \texttt{p} (producing a set of outputs) and then

run the same input with \texttt{q}. The result should be then just the union

of both sets, which is the operation \texttt{++} in Scala.

This parser combinator already allows us to construct a parser that either 

a character \texttt{a} or \texttt{b}, as

\begin{center}

\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]

new AltParser(CharParser('a'), CharParser('b'))

\end{lstlisting}

\end{center}

\noindent

Scala allows us to introduce some more readable shorthand notation for this, like \texttt{'a' || 'b'}. 

We can call this parser combinator with the strings

\begin{center}

\begin{tabular}{rcl}

input string & & output\medskip\\

\texttt{\Grid{ac}} & $\rightarrow$ & $\left\{(\texttt{\Grid{a}}, \texttt{\Grid{c}})\right\}$\\

\texttt{\Grid{bc}} & $\rightarrow$ & $\left\{(\texttt{\Grid{b}}, \texttt{\Grid{c}})\right\}$\\

\texttt{\Grid{cc}} & $\rightarrow$ & $\varnothing$

\end{tabular}

\end{center}

\noindent

We receive in the first two cases a successful output (that is a non-empty set).

A bit more interesting is the \emph{sequence parser combinator} implemented in

Scala as follows:

\begin{center}

\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]

class SeqParser[I, T, S]

       (p: => Parser[I, T], 

        q: => Parser[I, S]) extends Parser[I, (T, S)] {

  def parse(sb: I) = 

    for ((head1, tail1) <- p.parse(sb); 

         (head2, tail2) <- q.parse(tail1)) 

            yield ((head1, head2), tail2)

}

\end{lstlisting}

\end{center}

\noindent

This parser takes as input two parsers, \texttt{p} and \texttt{q}. It implements \texttt{parse} 

as follows: let first run the parser \texttt{p} on the input producing a set of pairs (\texttt{head1}, \texttt{tail1}).

The \texttt{tail1} stands for the unprocessed parts left over by \texttt{p}. 

Let \texttt{q} run on these unprocessed parts

producing again a set of pairs. The output of the sequence parser combinator is then a set

containing pairs where the first components are again pairs, namely what the first parser could parse

together with what the second parser could parse; the second component is the unprocessed

part left over after running the second parser \texttt{q}. Therefore the input type of

the sequence parser combinator is as usual \texttt{I}, but the output type is

\begin{center}

\texttt{Set[((T, S), I)]}

\end{center}

Scala allows us to provide some

shorthand notation for the sequence parser combinator. So we can write for 

example \texttt{'a'  $\sim$ 'b'}, which is the

parser combinator that first consumes the character \texttt{a} from a string and then \texttt{b}.

Calling this parser combinator with the strings

\begin{center}

\begin{tabular}{rcl}

input string & & output\medskip\\

\texttt{\Grid{abc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{a}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\

\texttt{\Grid{bac}} & $\rightarrow$ & $\varnothing$\\

\texttt{\Grid{ccc}} & $\rightarrow$ & $\varnothing$

\end{tabular}

\end{center}

\noindent

A slightly more complicated parser is \texttt{('a'  || 'b') $\sim$ 'b'} which parses as first character either

an \texttt{a} or \texttt{b} followed by a \texttt{b}. This parser produces the following results.

\begin{center}

\begin{tabular}{rcl}

input string & & output\medskip\\

\texttt{\Grid{abc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{a}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\

\texttt{\Grid{bbc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{b}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\

\texttt{\Grid{aac}} & $\rightarrow$ & $\varnothing$

\end{tabular}

\end{center}

Note carefully that constructing the parser \texttt{'a' || ('a' $\sim$ 'b')} will result in a tying error.

The first parser has as output type a single character (recall the type of \texttt{CharParser}),

but the second parser produces a pair of characters as output. The alternative parser is however

required to have both component parsers to have the same type. We will see later how we can 

build this parser without the typing error.